User talk:Allam948736
On last digits of tetrations Could you please give us a source of the last digits of tetrations? If they are your own results, could you tell us explicit algorithms which gave the results? Since the original description to compute the last digits in the article was wrong, I am afraid that you added wrong estimations following the article. Although I reverted your contributions, please rewrite them if you have sources available. Thanks. p-adic 05:01, December 18, 2019 (UTC) I still don't really get why the recursive modular exponentiation method shouldn't work in base 10. I have verified that it indeed doesn't work in most other numeral bases (such as 7, 11, 13, or 19), but using Mathematica, PowerMod5597484987, 10^10 gives the same result (6100739387) as PowerMod7625597484987, 10^10, and so the last digits of 3↑↑4 are indeed ...6100739387. Similarly, 3 to the power of any number ending in ...6100739387 ends in ...9660355387, thus those are indeed the last 10 digits of 3↑↑5. I could keep going and show that 3↑↑↑3, 3↑↑↑↑3, and even Graham's number do indeed end in ...2464195387. Allam948736 (talk) 19:53, December 30, 2019 (UTC) On first and last digits As I wrote above, you are adding your own results to many articles without external sources including a reproducible description on how you computed them. Your own site is not an evidence of your own results, because you have no actual written method to generate them. Please notice that even if you are certain that your results are completely correct, what you are doing is the same as adding random digits to articles for others because you lack descriptions of reproducible methods. If you do not add valid external sources, then I will start to delete all your contributions. p-adic 03:14, December 27, 2019 (UTC) I calculated the first digits of 3^3^100 by simply taking the log10 of 3 and multiplying by 3^100, and then taking 10 to the power of the fractional part, and the last digits by using simple modular exponentiation techniques. Which external sources are valid, exactly? Do I just have to add a calculation method to my own site's article, or do I have to search the Web for another source (which there may not be in some cases)? Allam948736 (talk) 03:31, December 27, 2019 (UTC) : Thank you for adding the explanations for \(2^{2^{100}}\). It is sufficient for me. Also, your site is valid for me, as long as you explicitly declare how you computed (in the article or your site). On the other hand, for example, something like "the last digits of tetration can be computed by modular exponentiation" is a bad explanation, because they are not easily computed in that way for higher torwers of exponentiations unlike lower towers such as \(2^{2^{100}}\) and \(3^{3^{100}}\). Could you tell me how you computed the last digits of tetration? As I commented above, your description here is not sufficient for others. : p-adic 04:12, December 27, 2019 (UTC) @p-bot This is not really related to the topic at hand, but I think you're being a tad too insistent Many artciles for numbers contain approximations in various other notations. Most of these approximations are not cited anywhere and there is nothing on the page that justifies the approximation. Do you think all those approximations should be deleted because of "no source"? Username5243 (talk) 23:49, December 30, 2019 (UTC) : Usually I do not care about such approximations, because if there is a wrong information, then we can ask the original authors the reasons. But if nobody can explain them in a reasonable way, then we should delete them. I am not requiring a full proof, but just requiring an explanation. : Allam's results are such ones. I asked Allam the precise methods so many times, but somewhy he or she hesitated to fix an answer. Moreover, Allam tried to shift goalposts when I pointed out incorrect logic in his or her "explanation". Then isn't it good for us to doubt that Allam executed a meaningful algorithm? : Also, I added counterexamples for the wrong methods in the original articles on Graham's number and tetration, which Allam believed to be correct. What should I do in addition? Are you requiring me to write a proof that Allam's secret wrong method occasionally outputs correct answers? : p-adic 00:45, December 31, 2019 (UTC) : : It was fine for the first digits of exponential factorials (where the last digits were recently removed). Also, why were the first and last digits of triangoogol and triangrolplex removed? Those were included in the source given (which is my site's article where I defined the googolisms). For the last time, the first digits are always computed using logarithms, and the last digits using modular arithmetic. ''' And no, I didn't "hesitate", because why would I keep a method for calculating numbers of all things secret? Also, those "counterexamples" themselves contain incorrect logic. First of all, the recursive modular exponentiation method in base 10 fails for just one digit (ex. 2^...6 = ...4 if ...6 is preceded by an even digit, but 2^...6 = ...6 if the first ...6 is preceded by an odd digit), but works for more digits (ex. 2^...36 = ...36 regardless of the digits before ...36 in the exponent). Second, "3^g64 - g64 is divisible by 10^x for any x" implies that we thought the recursive modular exponentiation method (that is, take a^b mod 10^d, then substitute that result for b, then repeat) would continue to work forever (which it doesn't, but in the case of Graham's number the number of digits you would have to go to get to that point is itself extremely large). Allam948736 (talk) 01:10, December 31, 2019 (UTC) :: > triangoogol and triangrolplex :: It is because the source was referred to at the first sentence. If the computations are also mentioned in the sources, i.e. your website, then it is my mistake. I will revert the two articles, and add the references also at the end of those computations. :: > For the last time, the first digits are always computed using logarithms, and the last digits using modular arithmetic. :: As I wrote above, "moduar exponentiation" or something like that is too ambiguous. For example, the wrong description in the article is also a modular exponentiation. Therefore I am requiring to add precise explanations of the methods which you used. :: > And no, I didn't "hesitate", because why would I keep a method for calculating numbers of all things secret? :: Then why haven't you ever written down explicit methods? I said that "modular exponentiation" is not a full reproducible description, but you have never described further details. Please add precise methods to each article. :: > Also, those "counterexamples" themselves contain incorrect logic. :: As I have already pointed out, your statement that my logic is incorrect is incorrect. Should I repeat the explanation? I showed a counterexample for the method in the original article, which has no restriction on the length of digits as it says "The last d digits of {}^y x in base b is defined by the following recursive formula:". I am not talking about your '''updated statement after you read my counterexamples. :: Also, even if you persist to state that it works for more digits d, you are wrong. Say, consider (x,y,b) = (10^d,2,b). I guess that you will still update your statement by saying "No, I am always considering the case where x is coprime with b" or something like that. :: > Second, "3^g64 - g64 is divisible by 10^x for any x" implies that we thought the recursive modular exponentiation method :: You might think so. It does not mean that others made the same mistake. :: p-adic 01:34, December 31, 2019 (UTC)